# Formula Rearrange

• July 12th 2012, 02:55 AM
UberTyson
Formula Rearrange
Hi,

I am generally not too bad at rearranging formulas, but the result I have for the below differs from that which my colleague has calculated. Other opinion(s) would be very much appreciated.
If possible a couple of intermediate steps would be useful for my understanding.

The equation is below and I would like to solve for X

Attachment 24271

When I rearrange i get:

Attachment 24272

[that is a B by the root sign]

I would very much appreciate another interpretation.

Regards

(I hope this is the correct forum I did think about it!)
• July 12th 2012, 03:48 AM
emakarov
Re: Formula Rearrange
We have $1 + \left(\frac{x}{C}\right)^B=\frac{A-D}{y-D}$, from where $\left(\frac{x}{C}\right)^B=\frac{A-D}{y-D}-1$, $\frac{x}{C}=\sqrt[B]{\frac{A-D}{y-D}-1}$ and $x=C\sqrt[B]{\frac{A-D}{y-D}-1}=C\sqrt[B]{\frac{A-y}{y-D}}$.
• July 12th 2012, 04:47 AM
UberTyson
Re: Formula Rearrange
Thanks very much for the response.

Would it be possible to explain 2 of the steps?:

First step: how does "(y - D)" become the denominator as opposed to just "y"

Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)

i apologise if I am being a bit thick here.

Regards
• July 12th 2012, 04:57 AM
emakarov
Re: Formula Rearrange
Quote:

Originally Posted by UberTyson
First step: how does "(y - D)" become the denominator as opposed to just "y"

In your version, the denominator is also y - D. We start from $y=\frac{A-D}{1+\left(\frac{x}{C}\right)^B}+D$, subtract D from both sides to get $y - D = \frac{A-D}{1+\left(\frac{x}{C}\right)^B}$, from where $1+\left(\frac{x}{C}\right)^B = \frac{A-D}{y-D}$ (just like 5 = 10 / 2 is equivalent to 2 = 10 / 5).

Quote:

Originally Posted by UberTyson
Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)

It's not just (A - D)/(y - D), but (A - D)/(y - D) - 1. Represent 1 as (y - D) / (y - D); now you have the same denominator, and A - D - (y - D) = A - y.
• July 12th 2012, 05:05 AM
earboth
Re: Formula Rearrange
Quote:

Originally Posted by UberTyson
Thanks very much for the response.

Would it be possible to explain 2 of the steps?:

First step: how does "(y - D)" become the denominator as opposed to just "y"

Let's start from the beginning:

$y=\frac{A-D}{1+\left(\frac xC \right)^B} + D~\implies~y-D=\frac{A-D}{1+\left(\frac xC \right)^B}~\implies~1+\left(\frac xC \right)^B = \frac{A-D}{y-D}$

Quote:

Last step: How (A - D)/(y - D) becomes (A - y)/(y - D)

...
$\frac{A-D}{y-D}-1 = \frac{A-D}{y-D}-\frac{y-D}{y-D} = \frac{A-D-y+D}{y-D} = \frac{A-y}{y-D}$